Forum Geometricorum
Volume 14 (2014) 349–368.
FORUM GEOM
ISSN 1534-1178
On Two Triads of Triangles Associated With the
Perpendicular Bisectors of the Sides of a Triangle
Shao-Cheng Liu
Abstract. We discover some properties of the triangle centers related to the two
triads of triangles associated with the perpendicular bisectors of the sides of a
triangle.
1. Introduction
Given a triangle T := ABC, let the perpendicular bisector of the side BC
intersect the sidelines AC and AB at Bc and Cb respectively. Define Ca , Ac , Ab ,
Ba similarly. In this paper we study the two triads of triangles
(i) Ta := AAb Ac , Tb := Ba BBc , Tc := Ca Cb C (see Figure 1a) and
(ii) Ta := ABa Ca , Tb := Ab BCb , Tc := Ac Bc C (see Figure 1b).
Cb
A
A
Bc
C
Bc
B
H
O
B
Cb
AcA
O
Ab
C
B
Ba
Ca
Ac
Ab
C
Ba
Ca
Figure 1a. Ta and orthic triangle
Figure 1b. The triangles Ta , Tb , Tc
Homogeneous barycentric coordinates are used throughout this work. With the
usual notations in triangle geometry, a, b, c for the lengths of the sides BC, CA,
AB, and
b2 + c2 − a2
c2 + a2 − b2
, SB =
,
2
2
the points on the perpendicular bisectors are
SA =
SC =
a2 + b2 − c2
,
2
Ab = (0 : −SA + SB : SA + SB ), Ac = (0 : SC + SA : SC − SA );
Bc = (SB + SC : 0 : −SB + SC ), Ba = (SA − SB : 0 : SA + SB );
Ca = (−SC + SA : SC + SA : 0), Cb = (SB + SC : SB − SC : 0).
Publication Date: November 4, 2014. Communicating Editor: Paul Yiu.
350
S.-C. Liu
Denote by T the orthic triangle of ABC. Its vertices are
A = (0 : SC : SB ),
B = (SC : 0 : SA ),
C = (SB : SA : 0).
The sidelengths a , b , c of B C , C A , A B of the orthic triangle are given by
a2 =
a2 SAA
,
b2 c2
b2 =
b2 SBB
,
c 2 a2
c2 =
c2 SCC
.
a2 b2
From these,
a2 : b2 : c2 = a4 SAA : b4 SBB : c4 SCC .
Corresponding to SA , SB , SC , we have
1 2
(b + c2 − a2 ) =
2
1
:= (c2 + a2 − b2 ) =
SB
2
1
SC := (a2 + b2 − c2 ) =
2
:=
SA
SABC S 2 − SAA
·
,
a2 b2 c2
SA
SABC S 2 − SBB
·
,
a2 b2 c2
SB
SABC S 2 − SCC
·
,
a2 b2 c2
SC
and
: SB
: SC = SBC (S 2 − SAA ) : SCA (S 2 − SBB ) : SAB (S 2 − SCC ).
SA
Lemma 1. (a) The triangles AAb Ac , Ba BBc , and Ca Cb C are all similar to T .
(b) The triangles ABa Ca , Ab BCb , and Ac Bc C are all similar to T.
2. A common point of circumcircles
Theorem 2. The circumcircles of triangles in the two triads (Ta , Tb , Tc ) and
(Ta , Tb , Tc ) all contain the Euler reflection point
E=
b2
c2
a2
:
:
b2 − c2 c2 − a2 a2 − b2
of the circumcircle of T.
Proof. We compute the coordinates of E with respect to these triangles, and show
from these coordinates that E lies on the circumcircle of each. With respect to T,
E = ((SB +SC )(SC −SA )(SA −SB ) : (SB −SC )(SC +SA )(SA −SB ) : (SB −SC )(SC −SA )(SA +SB ))
with coordinate sum σ = a2 SAA + b2 SBB + c2 SCC − 6SABC .
The coordinates of E in triangle Ta = AAb Ac are
Two triads of triangles
351
Cb
A
Bc
E
O
B
Ac
Ab
C
Ba
Ca
Figure 2
Area(EAb Ac ) : area(AEAc ) : area(AAb E)
(SB + SC )(SC − SA )(SA − SB )
1
0
=
4σSBC 0
1
1 (SB + SC )(SC − SA )(SA − SB )
:
2σSC 0
1
1 0
:
2σSB (S + S )(S − S )(S − S )
B
C
C
A
A
B
(SB − SC )(SC + SA )(SA − SB )
−(SA − SB )
SC + SA
0
(SB − SC )(SC + SA )(SA − SB )
SC + SA
(SB
0
−(SA − SB )
− SC )(SC + SA )(SA − SB )
(SB − SC )(SC − SA )(SA + SB )
SA + SB
SC − SA
0
(SB − SC )(SC − SA )(SA + SB )
SC − SA
0
SA + SB
(SB − SC )(SC − SA )(SA + SB )
SA (SC − SA )(SA − SB )(SB + SC )2
SB (SB − SC )(SC − SA )(SC + SA ) SC (SA − SB )(SB − SC )(SA + SB )
=
:
:
2σSBC
σSC
σSB
=
SBB (SC + SA ) SCC (SA + SB )
SA (SB + SC )2
:
:
2(SB − SC )
SA − SB
SC − SA
=
SAA (SB + SC )2
SBB (SC + SA )2
SCC (SA + SB )2
:
:
2SA (SB − SC )
(SC + SA )(SA − SB ) (SA + SB )(SC − SA )
=
a2
b2
c2
:
:
2SA (SB − SC )
(SC + SA )(SA − SB ) (SA + SB )(SC − SA )
This is the isogonal conjugate (in triangle Ta = AAb Ac ) of the point
(2SA (SB − SC ) : (SC + SA )(SA − SB ) : (SA + SB )(SC − SA )),
which is an infinite point since the coordinate sum
2SA (SB − SC ) + (SC + SA )(SA − SB ) + (SA + SB )(SC − SA ) = 0.
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S.-C. Liu
This shows that E is on the circumcircle of triangle Ta . Similarly, E is also on the
circumcircles of the triangles Tb and Tc , with coordinates
and
a2
b2
c2
:
:
(SB + SC )(SA − SB ) 2SB (SC − SA ) (SA + SB )(SB − SC )
a2
b2
c2
:
:
(SB + SC )(SC − SA ) (SC + SA )(SB − SC ) 2SC (SA − SB )
respectively. A similar calculation shows that E is also on the circumcircles of
triangles Ta , Tb , Tc , with coordinates
a2
b2
c2
:
:
,
2SA (SB − SC ) (SC + SA )(SA − SB ) (SA + SB )(SC − SA )
a2
b2
c2
:
:
,
(SB + SC )(SA − SB ) 2SB (SC − SA ) (SA + SB )(SB − SC )
b2
c2
a2
:
:
(SB + SC )(SC − SA ) (SC + SA )(SB − SC ) 2SC (SA − SB )
respectively in these triangles.
3. Counterparts of a point in the triad Ta , Tb , Tc
Let P be a point with homogeneous barycentric coordinates (x : y : z) with
respect to triangle T = ABC. The counterparts of P in the triangles Ta , Tb , Tc
are the points AP , BP , CP which have the same coordinates (x : y : z) in these
triangles. In homogeneous barycentric coordinates,
AP = (2SBC x : SC (−SA + SB )y + SB (SC + SA )z : SC (SA + SB )y + SB (SC − SA )z),
BP = (SA (SB + SC )z + SC (SA − SB )x : 2SCA y : SA (−SB + SC )z + SC (SA + SB )x),
CP = (SB (−SC + SA )x + SA (SB + SC )y : SB (SC + SA )x + SA (SB − SC )y : 2SAB z)
Denote by T(P ) the triangle AP BP CP . Basic properties of T(P ) can be found
in [1].
Theorem 3 (Bui). The triangle T(P ) is
(a) oppositely similar to T,
(b) orthologic to T,
(c) perspective with T if and only if P lies on the Jerabek hyperbola. In this case,
the perspector traverses the Euler line.
If P = (x : y : z), the perpendiculars from A to BP CP , B to CP AP , C to
AP BP concur at
1
: ··· : ··· ,
Q=
−b2 c2 SBC x + c2 SA SCC y + b2 SA SBB z
which lies on the circumcircle of T (see Figure 3).
Two triads of triangles
353
Cb
Q
A
CP
Bc
Q
OP
P
O
AP
BP
B
Ac
Ba
C
Ab
Ca
Figure 3.
On the other hand, the perpendiculars from AP to BC, BP to CA, CP to AB
concur at
Q = (SBC (a2 (SAA − SBC ) − SA (SBB + SCC )x + c2 a2 SA SCC y + a2 b2 SA SBB z
: · · · : · · · ).
According to [1], the coordinates of Q with respect to T(P ) are the same as
those of Q with respect to T. It follows that Q is a point on the circumcircle of
T(P ).
Proposition 4. The triangle T(P ) has orthocenter O.
Proof. The sum of the coordinates of AP given at the beginning of the present
section is SBC (x + y + z); similarly for BP and CP . The orthocenter of T(P ) has
coordinates
(2SBC x, SC (−SA + SB )y + SB (SC + SA )z, SC (SA + SB )y + SB (SC − SA )z),
+ (SA (SB + SC )z + SC (SA − SB )x, 2SCA y, SA (−SB + SC )z + SC (SA + SB )x),
+ (SB (−SC + SA )x + SA (SB + SC )y, SB (SC + SA )x + SA (SB − SC )y, 2SAB z)
= (x + y + z)(SA (SB + SC ), SB (SC + SA ), SC (SA + SB )).
This is the circumcenter O of T.
Theorem 5. The points AP , BP , CP and P are concyclic.
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S.-C. Liu
Proof. The coordinates of P with respect to T(P ) = AP BP CP are
Area(P BP CP ) : area(AP P CP ) : area(AP BP P )
=
(b2 SB x − a2 SA y)(a2 SA z − c2 SC x) (c2 SC y − b2 SB z)(b2 SB x − a2 SA y)
:
4SABC SA (x + y + z)2
4SABC SB (x + y + z)2
:
=
=
a2 S
A
y
b2 SB
(a2 SA z − c2 SC x)(c2 SC y − b2 SB z)
4SABC SC (x + y + z)2
(c2 S
a2
−
a2
b2
c2
: 2
: 2
2
2
2
2
b SB (a SA z − c SC x) c SC (b SB x − a2 SA y)
C y − b SB z)
z
c2 SC
:
z
c2 SC
b2
−
x
a2 S A
:
x
a2 S A
c2
−
y
b2 SB
.
This shows that P is the isogonal conjugate (in triangle T(P )) of an infinite point.
It is a point on the circumcircle of T(P ).
Since T(P ) is similar to T, its circumcenter is the point
a2 SA · AP + b2 SB · BP + c2 SC · CP
2S 2
in absolute barycentric coordinates. In homogeneous barycentric coordinates with
respect to T,
OP :=
OP = (SBC (4S 2 · SA − a2 b2 c2 )x + c2 a2 SCCA y + a2 b2 SABB z
: b2 c2 SBCC x + SCA (4S 2 · SB − a2 b2 c2 )y + a2 b2 SAAB z
: b2 c2 SBBC x + c2 a2 SCAA y + SAB (4S 2 · SC − a2 b2 c2 )z).
4. Counterparts of a point in the triad Ta , Tb , Tc
For P = (x : y : z) with respect to T, we also consider its counterparts AP ,
CP in the triangles Ta , Tb , Tc . These are the points
BP ,
AP = (2SA x − (a2 − b2 )y − (a2 − c2 )z : b2 z : c2 y),
BP = (a2 z : 2SB y − (b2 − c2 )z − (b2 − a2 )x : c2 x),
CP = (a2 y : b2 x : 2SC z − (c2 − a2 )x − (c2 − b2 )y).
Denote by T (P ) the triangle AP BP CP .
Proposition 6. For P = (x : y : z), the triangle T (P ) is
(a) oppositely similar to the orthic triangle T ,
(b) perspective with T at the isogonal conjugate of P in T, namely,
2
b2 c2
a
∗
:
:
.
P =
x
y
z
(c) orthologic to T if and only if P lies on the Euler line. In this case,
(i) the perpendiculars from A, B, C to BP CP , CP AP , AP BP are concurrent at
SB
SC
SA
,
:
:
X(265) =
S 2 − 3SAA S 2 − 3SBB S 2 − 3SCC
Two triads of triangles
355
Cb
A
Bc
CP
P
B
AP
P∗
BP
Ba
Ac
C
Ab
Ca
Figure 4.
(ii) if OP : P H = t : 1 − t, the perpendiculars from AP , BP , CP to BC, CA,
AB are concurrent at Q where OQ : QH = 1 − t : t.
Remark. X(265) is the reflection conjugate of O. Equivalently, it is the reflection
of O in the Jerabek center X(125).
Proposition 7. The points AP , BP , CP , P are concyclic if and only if P lies on the
circumconic which is the isogonal conjugate (in T) of the perpendicular bisector
of the segment OH.
Proof. With respect to T (P ), the point P has coordinates
x : y : z = Area(P BP CP ) : area(AP P CP ) : area(AP BP P )
=
2a2 SA x2 + a4 yz − (S 2 + 2SBC − SCC )zx − (S 2 + 2SBC − SBB )xy
−4SBC (x + y + z)2
:
2b2 SB y 2 + b4 zx − (S 2 + 2SCA − SAA )xy − (S 2 + 2SCA − SCC )yz
−4SCA (x + y + z)2
:
2c2 SC z 2 + c4 xy − (S 2 + 2SAB − SBB )yz − (S 2 + 2SAB − SAA )zx
−4SAB (x + y + z)2
= SA (2a2 SA x2 + a4 yz − (S 2 + 2SBC − SCC )zx − (S 2 + 2SBC − SBB )xy)
: SB (2b2 SB y 2 + b4 zx − (S 2 + 2SCA − SAA )xy − (S 2 + 2SCA − SCC )yz)
: SC (2c2 SC z 2 + c4 xy − (S 2 + 2SAB − SBB )yz − (S 2 + 2SAB − SAA )zx)
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S.-C. Liu
The point P lies on the circumcircle of T (P ) (which is similar to T ) if and only
if
0 = a2 y z + b2 z x + c2 x y = a4 SAA y z + b4 SBB z x + c4 SCC x y ⎛
⎞
= 2SABC · S 2 (x + y + z)2 |HP |2 ⎝
a4 (S 2 − 3SAA )yz ⎠ .
cyclic
There are two possibilities.
(i) |HP | = 0 =⇒ P = H. In this
case, 4AH 2= BH = CH = O.
(ii) P lies on the circumconic cyclic a (S − 3SAA )yz = 0, which is the
isogonal conjugate (in T) of the perpendicular bisector of the segment OH.
5. Triangle centers of Ta , Tb , Tc and Ta , Tb , Tc
Consider the circumcenter O = (a2 SA : b2 SB : c2 SC ). Note that AO , BO , CO
are not the circumcenters of the triangles Ta , Tb , Tc respectively. Indeed, these
three points coincide with O: AO = BO = CO = O. Instead, the circumcenters
of triangles Ta , Tb , Tc are the points
Oa = (a4 (S 2 − SAA ) : b2 (2S 2 · SB − SC (S 2 − SBB )) : c2 (2S 2 · SC − SB (S 2 − SCC ))),
Ob = (a2 (2S 2 · SA − SC (S 2 − SAA )) : b4 (S 2 − SBB ) : c2 (2S 2 · SC − SA (S 2 − SCC ))),
Oc = (a2 (2S 2 · SA − SB (S 2 − SAA )) : b2 (2S 2 · SB − SA (S 2 − SBB )) : c4 (S 2 − SCC )).
These form the vertices of T(P ) for
P = X(1147) = (a4 SA (S 2 − SAA ) : b4 SB (S 2 − SBB ) : c4 SC (S 2 − SCC )),
which, according to the E NCYCLOPEDIA OF TRIANGLE CENTERS [2], is the midpoint of O and X(155), the orthocenter of the tangential triangle (see Figure 5).
The three circumcenters are concyclic with X(1147) (see Theorem 5). The center of the circle containing them is X(156), the nine-point center of the tangential
triangle
More generally, let P be a triangle center of ABC, with coordinates expressed
in terms of a, b, c. The same triangle center Pa of Ta = AAb Ac is the point with
coordinates in which a, b, c are replaced by a , b , c respectively (likewise SA , SB ,
Two triads of triangles
357
A
Oc
X(156)
Hb
X(155)
X(1147)
G
Oa
O
H
Ha
X(68)
Ob
Hc
B
C
E
Figure 5.
, S , S respectively). For example, for orthocenters,
SC by SA
c
B
1
1
1
Ha =
: S : S
SA
B
C
SB
SC
SA
with respect to Ta
:
:
=
S 2 − SAA
S 2 − SBB
S 2 − SCC
SA
SB
(0, −SA + SB , SA + SB )
= 2
· (1, 0, 0) + 2
·
S − SAA
S − SBB
2SB
SC
(0, SC + SA , SC − SA )
+ 2
·
in absolute barycentric coordinates
S − SCC
2SC
= ((S 2 − SBB )(S 2 − SCC ) : a2 SC (S 2 − SAA ) : a2 SB (S 2 − SAA ))
with respect to T.
The orthocenters Ha , Hb , Hc are the vertices of T(Q) for
SB
SC
SA
.
:
:
Q = X(68) =
S 2 − SAA S 2 − SBB S 2 − SCC
The triangle center X(68) is the superior of X(1147). It lies on the circle containing the three orthocenters (see Figure 5 and Theorem 5). The circle Ha Hb Hc also
contains the orthocenter H.
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S.-C. Liu
The centroids Ga , Gb , Gc are the points
Ga = (2SBC : 2SBC + SA (SB − SC ) : 2SBC − SA (SB − SC )),
Gb = (2SCA − SB (SC − SA ) : 2SCA : 2SCA + SB (SC − SA )),
Gc = (2SAB + SC (SA − SB ) : 2SAB − SC (SA − SB ) : 2SAB ).
In this case, Ga Gb Gc = T(G).
On the other hand, since the triangles Ta = ABa Ca , Tb = Ab BCb , Tc =
Ca Cb C are similar to ABC, we have
Pa = AP ,
Pb = BP ,
Pc = Cp .
For example, the circumcenters are
Oa = AO = (3a2 SAA − SA (SB − SC )2 − a2 SBC : b2 c2 SC : b2 c2 SB ),
= (c2 a2 SC : 3b2 SBB − SB (SC − SA )2 − b2 SCA : c2 a2 SA ),
Ob = BO
Oc = CO
= (a2 b2 SB : a2 b2 SA : 3c2 SCC − SC (SA − SB )2 − c2 SAB ).
Ga = (2b2 + 2c2 − 3a2 : b2 : c2 ),
Gb = (a2 : 2c2 + 2a2 − 3b2 : c2 ),
Gc = (a2 : b2 : 2a2 + 2b2 − 3c2 ).
Ha = Hb = Hc = O.
6. Orthology with T and pedal triangles
6.1. The triangle Oa Ob Oc .
Proposition 8. The triangle Oa Ob Oc is orthologic to ABC.
(a) The perpendiculars from Oa to BC, Ob to CA, and Oc to AB concur at the
triangle center 1
Y(1) := a2 (SAA (SBB + SBC + SCC ) − SBB SCC ) : · · · : · · · .
(b) The perpendiculars from A to Ob Oc , B to Oc Oa , and C to Oa Ob concur at the
triangle center X(74) on the circumcircle of T.
Remark. The orthology center Y(1) lies on the circle Oa Ob Oc .
Proposition 9. The triangle Oa Ob Oc is orthologic with the pedal triangle of P if
and only if P lies on the Euler line. If P lies on this line,
(a) the perpendiculars from Oa , Ob , Oc to the corresponding sides of the pedal
triangle of P are concurrent at a point on the conic
a2 SA (SB − SC )yz − (x + y + z)(fa x + fb y + fc z) = 0,
16S 2 · SABC
cyclic
1The triangle center Y
(1) does not appear in the current edition of [2]. It has (6-9-13)-search
number −5.64011769173 · · · .
Two triads of triangles
359
A
X(74)
Oc
Y (1)
Oa
O
Ob
C
B
E
Figure 6
where
3
fa = b2 c2 (SB −SC )(a4 SA
+a2 SAA (SBB −3SBC +SCC )−(8SA +SB +SC )SBB SCC ),
and fb , fc are defined cyclically,
(b) the perpendiculars from the vertices of the pedal triangle of P to the corresponding sides of Oa Ob Oc are concurrent at a point on the line
cyclic
SA (SB − SC )
x = 0.
a2 SAA − SA (SB − SC )2 − a2 SBC
6.2. The triangle Ga Gb Gc .
Proposition 10. The triangle Ga Gb Gc is orthologic to ABC.
(a) The perpendiculars from Ga to BC, Gb to CA, and Gc to AB concur at the
triangle center 2
Y(2) := (a2 (SBB + SBC + SCC )SAA + SBB SCC (2SA − SB − SC ) : · · · : · · · ).
2The triangle center Y
(2) does not appear in the current edition of [2]. It has (6-9-13)-search
number −5.65228493146 · · · .
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S.-C. Liu
(b) The perpendiculars from A to Gb Gc , B to Gc Ga , and C to Ga Gb concur at the
triangle center
1
: ··· : ···
X(1294) =
SAA (SBB − SBC + SCC ) − SBB SCC
on the circumcircle.
A
Gc
G
X(1294)
Y(2)
O
Ga
Gb
C
B
E
Figure 7
Remark. The circle Ga Gb Gc contains the centroid G and the orthology center Y(2) .
Proposition 11. The triangle Ga Gb Gc is orthologic with the pedal triangle of P
if and only if P lies on the line
b2 c2 (SB − SC )(a2 SA − SBC )x = 0,
cyclic
2
2
2
which passes through O and X(64) = a2 SAa−SBC : b2 SBb−SCA : c2 SCc−SAB .
If P lies on this line,
(a) the perpendiculars from Ga , Gb , Gc to the corresponding sides of the pedal
triangle of P are concurrent at a point on the conic
6SABC
(SB − SC )(a2 SA − SBC )yz
cyclic
⎛
− (x + y + z) ⎝
cyclic
⎞
SA (SB − SC )(SAA (SBB − 3SBC + SCC ) − SBB SCC )x⎠ = 0.
Two triads of triangles
361
(b) the perpendiculars from the vertices of the pedal triangle of P to the corresponding sides of Ga Gb Gc are concurrent at a point on the line
SA (SB − SC )(a2 SA − SBC )
x = 0.
a2 SA − 2SBC
cyclic
6.3. The triangle Ha Hb Hc .
Proposition 12. The triangle Ha Hb Hc is orthologic to ABC.
(a) The perpendiculars from Ha to BC, Hb to CA, and Hc to AB concur at the
orthocenter H of T.
(b) The perpendiculars from A to Hb Hc , B to Hc Ha , and C to Ha Hb concur at
1
: ··· : ···
X(1300) =
SA (a2 (SAA − SBC ) − SA (SB − SC )2 )
on the circumcircle of T.
A
X(1300)
Hb
O
H
Ha
Hc
C
B
E
Figure 8
Remark. X(1300) is the second intersection of the circumcircle with the line EH.
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S.-C. Liu
Proposition 13. The triangle Ha Hb Hc is orthologic with the pedal triangle of P
if and only if P lies on the line
SC − SA
SA − SB
SB − SC
x+
y+
z=0
2
2
a SA
b SB
c2 SC
joining the circumcenter O to X(394) = (a2 SAA : b2 SBB : c2 SCC ).
If P lies on this line,
(a) the perpendiculars from Ha , Hb , Hc to the corresponding sides of the pedal
triangle of P are concurrent at a point on the conic
⎛
⎞
SBC (SB − SC )yz + (x + y + z) ⎝
a2 SA (SB − SC )(S 2 − SAA )x⎠ = 0,
4S 2
cyclic
cyclic
with center at the nine-point center N ;
(b) the perpendiculars from the vertices of the pedal triangle of P to the corresponding sides of Ha Hb Hc are concurrent at a point on the line
SB − SC
x = 0.
4
2
SA (a SAA − a SA (SBB + SCC ) − SBC (SB − SC )2 )
cyclic
6.4. The triangle Oa Ob Oc .
Proposition 14. The triangle Oa Ob Oc is orthologic to ABC.
(a) The perpendiculars from Oa to BC, Ob to CA, and Oc to AB concur at the
orthocenter H of T.
(b) The perpendiculars from A to Ob Oc , B to Oc Oa , and C to Oa Ob concur at the
triangle center
SB
SC
SA
:
:
X(265) =
S 2 − 3SAA S 2 − 3SBB S 2 − 3SCC
(see Proposition 6).
Proposition 15. The triangle Oa Ob Oc is orthologic with the pedal triangle of P if
and only if P lies on the Euler line. If P lies on this line,
(a) the perpendiculars from Oa , Ob , Oc to the corresponding sides of the pedal
triangle of P are concurrent at a point on the conic
a2 SA (SB − SC )yz
4S 2
cyclic
2 2 2
+ a b c (x + y + z)(SA (SB − SC )x + SB (SC − SA )y + SC (SA − SB )z) = 0,
with center N ,
(b) the perpendiculars from the vertices of the pedal triangle of P to the corresponding sides of Oa Ob Oc are concurrent at a point on the line
SB − SC
x = 0.
a2 SA − 2SBC
cyclic
Two triads of triangles
363
A
Ob
X(265)
O
Oc
B
E
C
Oa
Figure 9
6.5. The triangle Ga Gb Gc .
Proposition 16. The triangle Ga Gb Gc is orthologic to ABC.
(a) The perpendiculars from Ga to BC, Gb to CA, and Gc to AB concur at the
triangle center
X(381) = a2 SA + 4SBC : b2 SB + 4SCA : c2 SC + 4SAB
on the Euler line.
(b) The perpendiculars from A to Gb Gc , B to Gc Ga , and C to Ga Gb concur at the
triangle center
SB
SC
SA
:
:
X(265) =
S 2 − 3SAA S 2 − 3SBB S 2 − 3SCC
(see Proposition 6).
Proposition 17. The triangle Ga Gb Gc is orthologic with the pedal triangle of P
if and only if P lies on the Euler line. If P lies on this line,
(a) the perpendiculars from Ga , Gb , Gc to the corresponding sides of the pedal
triangle of P are concurrent at a point on the conic
⎛
⎞
6
a2 SA (SB − SC )yz + (x + y + z) ⎝
b2 c2 (SB − SC )x⎠ = 0
cyclic
cyclic
with center at the centroid G,
(b) the perpendiculars from the vertices of the pedal triangle of P to the corresponding sides of Ga Gb Gc are concurrent at a point on the line
SB − SC
x = 0.
a2 SA − 2SBC
cyclic
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S.-C. Liu
A
X(265)
Gb
X(381)
GO
Ga
Gc
B
C
E
Figure 10
7. The triangles T(P ) and T(Q)
Proposition 18. The triangles T(P ) and T(Q) are perspective if and only if the
line P Q contains the circumcenter O. In this case, the triangles are homothetic at
O.
Proof. Let P = (x : y : z) and Q = (u : v : w). The line containing AP and AQ
has equation
a2 SA (wy − vz)X + (c2 SC (vx − uy) − SB (SC − SA )(uz − wx))Y
+ (b2 SB (uz − wx) + SC (SA − SB )(vx − uy))Z = 0.
Similarly, we have the equations of the lines BP BQ and CP CQ . The three lines
are concurrent if and only if f · g = 0, where
f : = (c2 SC v − b2 SB w)x + (a2 SA w − c2 SC u)y + (b2 SB u − a2 SA v)z,
⎞
⎛
a4 SAA yz ⎠
g : = (u + v + w)2 ⎝
⎛
− (x + y + z) ⎝
cyclic
⎞
((c2 SC v + b2 SB w)2 − 4SABC · SA vw)x⎠ .
cyclic
Note that
(i) the equation f = 0 represents the line OQ;
(ii) the equation g = 0 represents a conic with center Q. Since the conic also
contains Q and intersects the sidelines at imaginary points, its represents only the
point Q.
From these we conclude that the triangles T(P ) and T(Q) are perspective if and
only if O, P , Q are collinear. Each of the lines AP AQ , BP BQ , CP CQ contains O.
The perspector is O.
Two triads of triangles
365
Theorem 19. The triangle bounded by the lines AP AQ , BP BQ , CP CQ has circumcenter O.
Proof. Since O is the orthocenter of the triangle T(P ), the circles OBP CP , OCP AP ,
and OAP BP have equal radii. Note that O is also the incenter of each of triangles
Ta , Tb , Tc , and these triangles are all similar to T . Therefore, ∠OAP AQ =
∠OBP BQ = ∠OCP CQ . Let A B C be the triangle bounded by the lines
AP AQ , BP BQ and CP CQ . Applying the law of sines to triangles OA CP , OB AP ,
OC BP , we conclude that OA = OB = OC .
Proposition 20. The triangles T (P ) and T (Q) are perspective if and only if Q
lies on the line HP . If the condition is satisfied, the triangles are homothetic at O.
8. The triangles T(P ) and T (P )
Theorem 21. The triangles T(P ) and T (P ) are perspective if and only if P lies
on
(a) the Jerabek
hyperbola or
(b) the line cyclic SA (SBx−SC ) = 0.
Remarks. (1) If P lies on the Jerabek hyperbola, the perspector is the circumcenter
O.
(2) The line in (b) contains the Jerabek center X(125) and X(122) (also X(684),
X(1650), X(2972)). In this case the lines AP AP , BP BP , CP CP are parallel.
Theorem 22. Let P be a point with coordinates (x : y : z) in T.
(a) The circumcenter of T(P ) has coordinates (x : y : z) in triangle Oa Ob Oc =
C .
T (O) = AO BO
O
(b) The circumcenter of T (P ) has coordinates (x : y : z) in Oa Ob Oc , which is
T(Q) for
Q = X(1147) = (a4 SA (S 2 − SAA ) : b4 SB (S 2 − SBB ) : c4 SC (S 2 − SCC )).
Proof. (a) The point with coordinates (x : y : z) with respect to Oa Ob Oc is
O = (SBC (4S 2 · SA − a2 b2 c2 )x + c2 a2 SCCA y + a2 b2 SABB z
: b2 c2 SBCC x + SCA (4S 2 · SB − a2 b2 c2 )y + a2 b2 SAAB z
: b2 c2 SBBC x + c2 a2 SCAA y + SAB (4S 2 · SC − a2 b2 c2 )z).
Its square distance from AP is
a2 b2 c2 Q(x, y, z)
,
16S 2 · (SABC )2 (x + y + z)2
where
Q(x, y, z) = b2 c2 SBB SCC x2 + c2 a2 SCC SAA y 2 + a2 b2 SAA SBB z 2
− 2SABC (a2 SAA yz + b2 SBB zx + c2 SCC xy).
The symmetry of Q in SA , SB , SC and x, y, z shows that this is the same if AP
is replaced by BP or CP . The point O therefore is the circumcenter of AP BP CP .
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S.-C. Liu
Theorem 23. (a) Triangles Oa Ob Oc and Oa Ob Oc are perspective at the circumcenter O of triangle ABC.
(b) The six circumcenters Oa , Ob , Oc , Oa , Ob , Oc are concyclic. The center of the
circle containing them is X(156), the nine-point center of the tangential triangle.
Cb
A
Bc
E
Ob
Oa
Oc
B
O
Oc
Ob
Ac
Ab
Oa
C
Ba
Ca
Figure 11
Proof. With respect to Oa Ob Oc ,
−a2
b2
c2
Oa = (−a2 SBC : SC (a2 SA + 2SBC ) : SB (a2 SA + 2SBC )) =
:
:
,
a2 SA + 2SBC b2 SB c2 SC
2
a
−b2
c2
: 2
: 2
Ob = (SC (b2 SB + 2SCA ) : −b2 SBC : SA (b2 SB + 2SCA )) =
,
2
a SA b SB + 2SCA c SC
2
a
b2
−c2
:
:
.
Oc = (SB (c2 SC + 2SAB ) : SA (c2 SC + 2SAB ) : −c2 SBC ) =
a2 SA b2 SB c2 SC + 2SAB
These expressions show that
(a) triangles Oa Ob Oc and Oa Ob Oc are perspective at the orthocenter of Oa Ob Oc ,
(b) Oa , Ob , Oc are on the the circumcircle of Oa Ob Oc .
The orthocenter of Oa Ob Oc is the circumcenter O. The circumcenter is X(156),
which is the nine-point center of the tangential triangle.
Proposition 24. The triangle Oa Ob Oc is perspective with the inferior triangle of
the tangential triangle at a point on its circumcircle.
Proof. With respect to Oa Ob Oc , the midpoint of the A-side of the tangential triangle has coordinates
(a2 SAA (2SBB − 7SBC + 2SCC ) + SABC (3SBB − 2SBC + 3SCC ) + a2 SBB SCC )
: a2 b2 SA (c2 SC − 2SAB ) : c2 a2 SA (b2 SB − 2SCA ));
Two triads of triangles
367
similarly for the midpoints of the other two sides of the tangential triangle. From
these coordinates, it is clear that the medial triangle of the tangential triangle is
perspective to Oa Ob Oc at the point with coordinates
b2
c2
a2
.
:
:
a2 SA − 2SBC b2 SB − 2SCA c2 SC − 2SAB
Since triangle Oa Ob Oc is similar to T, this perspector is a point on the circumcircle
of Oa Ob Oc . It is the isogonal conjugate (with respect to Oa Ob Oc ) of the infinite
point of the Euler line of the triangle.
Remark. With respect to T, this perspector has coordinates
(a2 (SAA (SBB + SBC + SCC ) − SBB SCC ) : · · · : · · · ).
See Proposition 8.
Proposition 25. The triangle Oa Ob Oc is perspective with the tangential triangle
at
3
+(5SBB +6SBC +5SCC )SAA +9a2 SABC +4SBB SCC ) : · · · : · · · ).
X(195) = (a2 (−3a2 SA
Remark. X(195) is the circumcenter of the triangle of reflections (see [3]).
9. A family of circumcircles of T(P ) containing the Euler reflection point
Proposition 26. Let P = (x : y : z). The circumcircle of T(P ) = AP BP CP
contains the Euler reflection point E if and only if P lies on the conic
a4 SAA yz + b4 SBB zx + c4 SCC xy = 0.
Proof. The coordinates of AP , BP , CP are given at the beginning of §3. Computing the coordinates of E with respect to triangle AP BP CP , we have
x : y : z ⎛ ⎛
⎞
⎞
S
2S
x
y
z
2S
BC
AAC
AAB
⎠
+
= SBC ⎝μ ⎝
a4 SAA yz ⎠ + τ (x + y + z)
+ 2
SB − SC
b (SA − SB ) c2 (SC − SA )
cyclic
⎛ ⎛
⎞
⎞
2S
S
x
y
z
2S
BBC
CA
ABB
⎠
+
: SCA ⎝μ ⎝
a4 SAA yz ⎠ + τ (x + y + z)
+ 2
a2 (SA − SB ) SC − SA
c (SB − SC )
cyclic
⎛ ⎛
⎞
⎞
2S
2S
S
x
y
z
BCC
CCA
AB
⎠
+
+
a4 SAA yz ⎠ + τ (x + y + z)
: SAB ⎝μ ⎝
a2 (SC − SA ) b2 (SB − SC ) SA − SB
cyclic
where
μ = a2 SBC + b2 SCA + c2 SAB − 6SABC ,
τ = (SBB − SCC )(SCC − SAA )(SAA − SBB ).
This is a point on the circumcircle of triangle AP BP CP (which is similar to ABC)
if and only if
(SB + SC )y z + (SC + SA )z x + (SA + SB )x y = 0.
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S.-C. Liu
⎛
This reduces to
μ2 · ⎝
⎞
a4 SAA yz ⎠ · G = 0,
cyclic
where
G=
(b2 c2 SBB SCC x2 − 2SABC · a2 SAA yz)
cyclic
⎛
= S2 ⎝
⎞
2
2
SBC
x2 − 2SCA SAB yz ⎠ + SABC
(x + y + z)2
cyclic
G = 0 is the equation of a conic with center O. Since G(O) = 0, we conclude
that G = 0 defines only the point O. Therefore,
the circle AP BP CP contains E if
and only if P lies on the circumconic cyclic a4 SAA yz.
Theorem 27. Let Q be a triangle center on the circumcircle. The circle Qa Qb Qc
contains the Euler reflection point E.
2
2
2
Proof. Let Q = (b2 −c2a)(a2 +t) : (c2 −a2b)(b2 +t) : (a2 −b2c)(c2 +t) be a triangle center
on the circumcircle. For
b2
c2
a2
:
:
P =
(b2 − c2 )(a2 + t) (c2 − a2 )(b2 + t) (a2 − b2 )(c2 + t)
we have Qa = AP , Qb = BP , Qc = CP . The circle Qa Qb Qc is the same as
AP BP CP . With respect to triangle Oa Ob Oc , the center of AP BP CP has coordinates given above. Therefore the locus of the center of Qa Qb Qc is the circle
Oa Ob Oc , which is the nine-point circle
of the tangential triangle of triangle ABC.
Note that P lies on the conic cyclic a4 SAA yz = 0. By Proposition 26, the
circle AP BP CP contains the Euler reflection point E.
Theorem 28. The circumcircle of T (P ) contains the Euler reflection point if and
only if P lies on the circumcircle.
References
[1] Q. T. Bui, On a triad of similar triangles associated with the perpendicular bisectors of the sides
of a triangle, Forum Geom., 10 (2010) 1–6.
[2] C. Kimberling, Encyclopedia of Triangle Centers, available at
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
[3] J. Torres, The triangle of reflections, Forum Geom., 14 (2014) 265–294.
[4] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,
2001; with corrections, 2013, available at
http://math.fau.edu/Yiu/Geometry.html
Shao-Cheng Liu: 2F., No.8, Alley 9, Lane 22, Wende Rd., 11475 Taipei, Taiwan
E-mail address: [email protected]